Some basic concepts of thermodynamics help us in understandingthe theory of metallurgical transformations. Gibb’s energy is the mostsignificant term.
The graphical representation of Gibb’s energy was first used by H.J.T.Ellingham. This provides a sound basis for considering the choice ofreducing agent in the reduction of oxides. This is known as Ellinghamdiagram. (For diagram refer answer number 3) Such diagrams help usin predicting the feasibility of thermal reduction of an ore. The criterionof feasibility is that at a given temperature Gibb’s energy of the reactionmust be negative.
Examples :
(i) Thermodynamics helps us to understand how coke reduces the oxideand why blast furnace is chosen. One of the main reduction steps in thisprocess is :
FeO(s)+C(s) –––––▶Fe(s∕l)+CO(g) .....(1)
It can be seen as a couple of two simpler reactions. In one, the reduction ofFeO is taking place and in the other, C is being oxidised to CO:
C(s)+

1

2

O2(g) –––––▶CO(g) [ΔG( C, CO) ] .....(3)
When both the reactions take place to yield the equation (1), the net Gibb’senergy change becomes :
ΔG( C, CO)+ΔG( FeO, Fe)=ΔrG .....(4)
Naturally, the resultant reaction will take place when the right hand sidein equation (3) is negative. In ∆G° vs T plot representing reaction 2, theplot goes upward and that representing the change C → CO (C, CO) goesdownward. At temperatures above 1073 K (approx), the C, CO line comesbelow the Fe, FeO line [∆G(C,CO)<∆G(Fe,FeO)]. So in this range, coke willbe reducing the FeO and will itself be oxidised to CO. In a similar waythe reduction of Fe3O4 and Fe2O3 at relatively lower temperatures by COcan be explained on the basis of lower lying points of intersection of theircurves with the CO,CO2 curve in Ellinghan diagram.
(ii) In the graph of ∆rG° vs T for formation of oxides, the Cu2O line isalmost at the top. So it is quite easy to reduce oxide ores of copper directlyto the metal by heating with coke (both the lines of C, CO and C, CO2are at much lower position in the graph particularly after 500 - 600 K).However most of the ores are sulphide and some may also contain iron.The sulphide ores are roasted/smelted to give oxides:
2 Cu2S+3 O2–––––▶2 Cu2O+2SO2
The oxide can then be easily reduced to metallic copper using coke.
Cu2O+C–––––▶2Cu+ CO